Home Solver RLC circuit in alternating current Physics 2 · Exercise 8 of 13
EM Induction Hard
RLC circuit in alternating current A series RLC circuit has R = 50 Ω , L = 0.2 H , C = 50 μF , powered at V = 220 V (RMS) , f = 60 Hz .\nCalculate: (a) X_L and X_C, (b) total impedance Z, (c) RMS current I, (d) resonance frequency.
Given data
R = 50 Ω L = 0.2 H C = 50×10⁻⁶ F V_rms = 220 V f = 60 Hz ω = 2πf ≈ 376.99 rad/s Review the theory: Induzione EM 1 Calculate the inductive reactance X_L = ωL (in Ω).
2 Calculate the capacitive reactance X_C = 1/(ωC) (in Ω).
3 Calculate the total impedance Z (in Ω).
4 Calculate the resonance frequency f₀ (in Hz) at which Z is minimum.
Calculate the inductive reactance X_L = ωL (in Ω).
X L = ω L = 2 π f L = 2 π × 60 × 0.2 = 2 π × 12 = 24 π ≈ 75.4 Ω X_L = \omega L = 2\pi f L = 2\pi \times 60 \times 0.2 = 2\pi \times 12 = 24\pi \approx \mathbf{75.4\,\Omega} X L = ω L = 2 π f L = 2 π × 60 × 0.2 = 2 π × 12 = 24 π ≈ 75.4 Ω . The inductive reactance is proportional to frequency and inductance.
Calculate the capacitive reactance X_C = 1/(ωC) (in Ω).
X C = 1 ω C = 1 2 π × 60 × 50 × 10 − 6 = 1 376.99 × 50 × 10 − 6 = 1 0.01885 ≈ 53.05 Ω X_C = \frac{1}{\omega C} = \frac{1}{2\pi \times 60 \times 50\times10^{-6}} = \frac{1}{376.99 \times 50\times10^{-6}} = \frac{1}{0.01885} \approx \mathbf{53.05\,\Omega} X C = ω C 1 = 2 π × 60 × 50 × 1 0 − 6 1 = 376.99 × 50 × 1 0 − 6 1 = 0.01885 1 ≈ 53.05 Ω . The capacitive reactance decreases with increasing frequency.
Calculate the total impedance Z (in Ω).
X L − X C = 75.4 − 53.05 = 22.35 Ω X_L - X_C = 75.4 - 53.05 = 22.35\,\Omega X L − X C = 75.4 − 53.05 = 22.35 Ω .
Z = R 2 + ( X L − X C ) 2 = 50 2 + 22.35 2 = 2500 + 499.5 = 2999.5 ≈ 54.77 Ω Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{50^2 + 22.35^2} = \sqrt{2500 + 499.5} = \sqrt{2999.5} \approx \mathbf{54.77\,\Omega} Z = R 2 + ( X L − X C ) 2 = 5 0 2 + 22.3 5 2 = 2500 + 499.5 = 2999.5 ≈ 54.77 Ω .
Calculate the resonance frequency f₀ (in Hz) at which Z is minimum.
f 0 = 1 2 π L C = 1 2 π 0.2 × 50 × 10 − 6 = 1 2 π 10 − 5 = 1 2 π × 0.003162 ≈ 50.3 H z f_0 = \frac{1}{2\pi\sqrt{LC}} = \frac{1}{2\pi\sqrt{0.2 \times 50\times10^{-6}}} = \frac{1}{2\pi\sqrt{10^{-5}}} = \frac{1}{2\pi \times 0.003162} \approx \mathbf{50.3\,Hz} f 0 = 2 π L C 1 = 2 π 0.2 × 50 × 1 0 − 6 1 = 2 π 1 0 − 5 1 = 2 π × 0.003162 1 ≈ 50.3 Hz . At resonance,
X L = X C X_L = X_C X L = X C and the impedance is purely resistive (
Z = R = 50 Ω Z = R = 50\,\Omega Z = R = 50 Ω ).
Result: X_L = 75.4 Ω , X_C = 53.1 Ω — Z = 54.8 Ω — I = 4.02 A — f₀ = 50.3 Hz .
